Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $r = \dfrac{-7}{3(4k - 3)} \div \dfrac{2}{28k^2 - 21k} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-7}{3(4k - 3)} \times \dfrac{28k^2 - 21k}{2} $ When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ -7 \times (28k^2 - 21k) } { 3(4k - 3) \times 2 } $ $ r = \dfrac {-7 \times 7k(4k - 3)} {2 \times 3(4k - 3)} $ $ r = \dfrac{-49k(4k - 3)}{6(4k - 3)} $ We can cancel the $4k - 3$ so long as $4k - 3 \neq 0$ Therefore $k \neq \dfrac{3}{4}$ $r = \dfrac{-49k \cancel{(4k - 3})}{6 \cancel{(4k - 3)}} = -\dfrac{49k}{6} $